Question

If first three terms of sequence $\frac{1}{16},a,b,\frac{1}{6}$ are in geometric series and last three terms are in harmonic series, then the value of $a$ and $b$ will be

• A. $a=-\frac{1}{4},b=1$
• B. $a=\frac{1}{12},b=\frac{1}{9}$
• C. $A$ and $B$ both are true
• D. None of these

Answer 1

The correct option is C

$A$ and $B$ both are true

Explanation for the correct option:

Step 1. Find the value of $a$ and $b$:

Given, $\frac{1}{16},a,b$ are in GP

$\Rightarrow$$a^2=\frac{b}{16}$ ….(1)

Also, $a,b,\frac{1}{6}$ are in HP

$\Rightarrow$$b=\frac{{\displaystyle \frac{2a}{6}}}{a+{\displaystyle \frac{1}{6}}}$

$\Rightarrow$$b=\frac{2a}{6a+1}$ ….(2)

Step 2. Substitute the value of $b$ in equation (1), we get

$a^2=\frac{2a}{16\left(6a+1\right)}$

$\Rightarrow$ $a^2=\frac{a}{8\left(6a+1\right)}$

$\Rightarrow$ $a=\frac{1}{8\left(6a+1\right)}$

$\Rightarrow$ $48a^2+8a=1$

$\Rightarrow$ $48a^2+8a-1=0$

$\Rightarrow$$(12a-1)(4a+1)=0$

$\mathbf{\therefore }\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{12}}\mathbf{,}\mathbf{}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}$

Step 3. Find the value of $b$ by putting the values of $a$:

When $a=\frac{1}{12}$,

$\begin{array}{rcl}b& =& \frac{2\left({\displaystyle \frac{1}{12}}\right)}{6\left(\frac{1}{12}\right)+1}\\ & =& \frac{2}{18}\\ & =& \frac{1}{9}\end{array}$

When $a=-\frac{1}{4}$,

$\begin{array}{rcl}b& =& \frac{2\left(-{\displaystyle \frac{1}{4}}\right)}{6\left(-\frac{1}{4}\right)+1}\\ & =& \frac{-2}{-2}\\ & =& 1\end{array}$

$\mathbf{\therefore }\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{9}}\mathbf{,}\mathbf{}\mathbf{1}$

Hence, Option ‘C’ is Correct.